CHAPTER 15 CHEMICAL EQUILIBRIUM 15.1

Equilibrium is a state in which no observable changes occur with time. There are two types of equilibria: physical and chemical. An example of a physical dynamic equilibrium is the vaporization and condensation of water in a closed container at a given temperature. A chemical example of a dynamic equilibrium is the decomposition and recombination of N2O4 and NO2 gases at a given temperature.

15.2

Equilibrium between two phases of the same substance is called physical equilibrium because the changes that occur are physical processes. When the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products no longer change with time, chemical equilibrium is reached. The equilibrium between solid and liquid during melting and the equilibrium between liquid and gas during boiling are examples of physical equilibria. The equilibrium between NO2 and N2O4 gases shown in Section 15.1 of the text and the equilibrium between CaCO3, CaO, and CO2 shown in Figure 15.3 of the text are examples of chemical equilibria.

15.3

Equilibrium processes are characterized by an equilibrium constant. The equilibrium constant provides information about the net direction of a reversible reaction and the concentrations of the equilibrium mixture.

15.4

Please see Figure15.1 of the text. Your graphs will look similar to these if NO2 = A and N2O4 = B. The differences between your graphs and those in Figure 15.1 will be that the increase or decrease in concentration of A will be steeper due to the 3:1 mole ratio in the balanced equation compared to the 2:1 mole ratio between NO2 and N2O4. Also, in your third graph, your starting concentration of A will be greater than B, which differs from what is shown in Figure 15.1 c.

15.5

Homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. The decompositions of N2O4 and PCl5 are examples. A reversible reaction involving reactants and products that are in different phases leads to a heterogeneous equilibrium. The decomposition of calcium carbonate (CaCO3) and the Mond process, the production of nickel tetracarbonyl [Ni(CO)4] from nickel and carbon monoxide gas, are examples.

15.6

Kc represents the equilibrium constant for an equilibrium process in which concentrations of reacting species are expressed in units of moles per liter. KP represents the equilibrium constant for a gas-phase equilibrium process in which concentrations are expressed in units of partial pressure.

15.7

(a)

(b)

Kc 

Kc 

[CO]2 [O2 ] [CO2 ]2

[O3 ]2 [O2 ]3

KP 

KP 

(c)

Kc 

[COCl2 ] [CO][Cl2 ]

KP 

(d)

Kc 

[CO][H 2 ] [H 2 O]

KP 

2 PCO  PO2 2 PCO

2

PO2

3

PO3

2

PCOCl2 PCO  PCl2

PCO  PH 2 PH 2O

350

15.8

15.9

CHAPTER 15: CHEMICAL EQUILIBRIUM

[H  ][HCOO  ] [HCOOH]

(e)

Kc 

(f)

K c  [O 2 ]

(a)

K P  PCO2 PH2O

(b)

2 K P  PSO PO 2

K P  PO2

2

[NH3 ]2

(a)

Kc 

(b)

Kc 

(c)

Kc 

[CO]2 [CO 2 ]

(d)

Kc 

[C6 H5 COO ][H  ] [C6 H5 COOH]

2

KP 

7

[NO2 ] [H 2 ] [SO2 ]2 3

[O2 ]

KP 

KP 

2 PSO

2 PNH

3

2 PNO P7 2 H2

2

PO3 2 2 PCO PCO2

Δn

15.10

KP = Kc(0.0821 T) . T is the temperature in Kelvin, 0.0821 is the gas constant, and Δn = moles of gaseous products  moles of gaseous reactants.

15.11

Kc 

[B] [A]

(1)

With Kc  10, products are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of B to be 10 times that of A at equilibrium. Choice (a) is the best choice with 10 B molecules and 1 A molecule.

(2)

With Kc  0.10, reactants are favored at equilibrium. Because the coefficients for both A and B are one, we expect the concentration of A to be 10 times that of B at equilibrium. Choice (d) is the best choice with 10 A molecules and 1 B molecule.

You can calculate Kc in each case without knowing the volume of the container because the mole ratio between A and B is the same. Volume will cancel from the Kc expression. Only moles of each component are needed to calculate Kc. 15.12

Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with coefficients of one in the balanced equation. (a)

The reaction, A  C ƒ AC has the largest equilibrium constant. Of the three diagrams, there is the most product present at equilibrium.

CHAPTER 15: CHEMICAL EQUILIBRIUM

(b)

15.13

The reaction, A  D ƒ AD has the smallest equilibrium constant. Of the three diagrams, there is the least amount of product present at equilibrium.

When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. K' 

15.14

351

1 1   2.40  1033  34 K 4.17  10

The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations into the equilibrium constant expression to calculate Kc. Step 1: Calculate the concentrations of the components in units of mol/L. The molarities can be calculated by simply dividing the number of moles by the volume of the flask. 2.50 mol  0.208 M 12.0 L

[H 2 ] 

[S2 ] 

1.35  105 mol  1.13  106 M 12.0 L

[H 2S] 

8.70 mol  0.725 M 12.0 L

Step 2: Once the molarities are known, Kc can be found by substituting the molarities into the equilibrium constant expression. Kc 

[H 2S]2 2

[H 2 ] [S2 ]



(0.725)2 6

2

(0.208) (1.13  10 )

 1.08  107

If you forget to convert moles to moles/liter, will you get a different answer? Under what circumstances will the two answers be the same? 15.15

Using Equation (15.5) of the text: KP  Kc(0.0821 T)

n

where, n  2  3  1 and T  (1273  273) K  1546 K 22

KP  (2.24  10 )(0.0821  1546) 15.16

1

 1.76  10

20

Strategy: The relationship between Kc and KP is given by Equation (15.5) of the text. What is the change in the number of moles of gases from reactant to product? Recall that n  moles of gaseous products  moles of gaseous reactants What unit of temperature should we use? Solution: The relationship between Kc and KP is given by Equation (15.5) of the text. KP  Kc(0.0821 T)

n

352

CHAPTER 15: CHEMICAL EQUILIBRIUM

Rearrange the equation relating KP and Kc, solving for Kc. Kc 

KP

(0.0821T ) n

Because T  575 K and n  3  2  1, we have: Kc 

15.17

KP

(0.0821T )n



5.0  104  1.1  105 (0.0821)(575 K)

We can write the equilibrium constant expression from the balanced equation and substitute in the pressures. KP 

2 PNO (0.050)2   0.051 PN2 PO2 (0.15)(0.33)

Do we need to know the temperature? 15.18

The equilibrium constant expressions are: (a)

Kc 

(b)

Kc 

[NH3 ]2 [N 2 ][H 2 ]3 [NH3 ] 3

1

[N 2 ] 2 [H 2 ] 2

Substituting the given equilibrium concentration gives: (a)

Kc 

(b)

Kc 

(0.25)2 (0.11)(1.91)3

 0.082

(0.25) 1

3

 0.29

(0.11) 2 (1.91) 2

Is there a relationship between the Kc values from parts (a) and (b)? 15.19

The equilibrium constant expression for the two forms of the equation are: Kc 

[I]2 and [I2 ]

K c' 

[I2 ] [I]2

The relationship between the two equilibrium constants is K c' 

1 1   2.6  104 Kc 3.8  105

KP can be found using Equation (15.5) of the text. KP  Kc'(0.0821 T)

n

4

 (2.6  10 )(0.0821  1000)

1

 3.2  10

2

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.20

353

Because pure solids do not enter into an equilibrium constant expression, we can calculate KP directly from the pressure that is due solely to CO2(g).

K P = PCO2  0.105 Now, we can convert KP to Kc using the following equation. KP  Kc(0.0821 T) KP

Kc 

(0.0821T ) n 0.105

Kc 

15.21

n

(0.0821  623)(1  0)

 2.05  103

We substitute the given pressures into the reaction quotient expression. QP 

PPCl3 PCl2 PPCl5



(0.223)(0.111)  0.140 (0.177)

The calculated value of QP is less than KP for this system. The system will change in a way to increase QP until it is equal to KP. To achieve this, the pressures of PCl3 and Cl2 must increase, and the pressure of PCl5 must decrease. Could you actually determine the final pressure of each gas? 15.22

Strategy: Because they are constant quantities, the concentrations of solids and liquids do not appear in the equilibrium constant expressions for heterogeneous systems. The total pressure at equilibrium that is given is due to both NH3 and CO2. Note that for every 1 atm of CO2 produced, 2 atm of NH3 will be produced due to the stoichiometry of the balanced equation. Using this ratio, we can calculate the partial pressures of NH3 and CO2 at equilibrium. Solution: The equilibrium constant expression for the reaction is 2 K P  PNH PCO2 3

The total pressure in the flask (0.363 atm) is a sum of the partial pressures of NH3 and CO2.

PT  PNH3  PCO2  0.363 atm Let the partial pressure of CO2  x. From the stoichiometry of the balanced equation, you should find that PNH3  2 PCO2 . Therefore, the partial pressure of NH3  2x. Substituting into the equation for total pressure gives:

PT  PNH3  PCO2  2 x  x  3x 3x  0.363 atm

x  PCO2  0.121 atm PNH3  2 x  0.242 atm

354

CHAPTER 15: CHEMICAL EQUILIBRIUM

Substitute the equilibrium pressures into the equilibrium constant expression to solve for KP. 2 K P  PNH PCO2  (0.242) 2 (0.121)  7.09  103 3

15.23

Of the original 1.05 moles of Br2, 1.20% has dissociated. The amount of Br2 dissociated in molar concentration is: 1.05 mol [Br2 ]  0.0120   0.0129 M 0.980 L Setting up a table:

ƒ Br2(g) 1.05 mol  1.07 M 0.980 L

Initial (M):

0.0129 1.06

Change (M): Equilibrium (M):

Kc 

15.24

2Br(g) 0 2(0.0129) 0.0258

[Br]2 (0.0258) 2   6.3  104 [Br2 ] 1.06

If the CO pressure at equilibrium is 0.497 atm, the balanced equation requires the chlorine pressure to have the same value. The initial pressure of phosgene gas can be found from the ideal gas equation. P 

nRT (3.00  102 mol)(0.0821 L  atm/mol  K)(800 K)   1.31 atm V (1.50 L)

Since there is a 1:1 mole ratio between phosgene and CO, the partial pressure of CO formed (0.497 atm) equals the partial pressure of phosgene reacted. The phosgene pressure at equilibrium is: Initial (atm): Change (atm): Equilibrium (atm):

CO(g)  0 0.497 0.497

Cl2(g) ƒ COCl2(g) 0 1.31 0.497 0.497 0.497 0.81

The value of KP is then found by substitution. KP 

15.25

PCOCl2 PCO PCl2



0.81 (0.497) 2

 3.3

Let x be the initial pressure of NOBr. Using the balanced equation, we can write expressions for the partial pressures at equilibrium. PNOBr  (1  0.34)x  0.66x PNO  0.34x

PBr2  0.17x The sum of these is the total pressure. We carry an additional significant figure throughout the calculations to minimize rounding errors. 0.66x  0.34x  0.17x  1.17x  0.25 atm x  0.214 atm

CHAPTER 15: CHEMICAL EQUILIBRIUM

The equilibrium pressures are then PNOBr  0.66(0.214)  0.141 atm PNO  0.34(0.214)  0.0728 atm

PBr2  0.17(0.214)  0.0364 atm We find KP by substitution. KP 

( PNO ) 2 PBr2 ( PNOBr )



2

(0.0728) 2 (0.0364) (0.141)

2

 9.7  103

The relationship between KP and Kc is given by KP  Kc(0.0821 T)

n

We find Kc (for this system n  1) Kc 

15.26

KP

( RT )n



KP 9.7  103   4.0  104 1 RT (0.0821  298)

In this problem, you are asked to calculate Kc. Step 1: Calculate the initial concentration of NOCl. We carry an extra significant figure throughout this calculation to minimize rounding errors. [NOCl]0 

2.50 mol  1.667 M 1.50 L

Step 2: Let's represent the change in concentration of NOCl as 2x. Setting up a table: Initial (M): Change (M): Equilibrium (M):

2NOCl(g) ƒ 2NO(g)  Cl2(g) 1.667 0 0 2x 2x x 1.667  2x 2x x

If 28.0 percent of the NOCl has dissociated at equilibrium, the amount reacted is: (0.280)(1.667 M)  0.4668 M In the table above, we have represented the amount of NOCl that reacts as 2x. Therefore, 2x  0.4668 M x  0.2334 M The equilibrium concentrations of NOCl, NO, and Cl2 are: [NOCl]  (1.67  2x)M  (1.667  0.4668)M  1.200 M [NO]  2x  0.4668 M [Cl2]  x  0.2334 M

355

356

CHAPTER 15: CHEMICAL EQUILIBRIUM

Step 3: The equilibrium constant Kc can be calculated by substituting the above concentrations into the equilibrium constant expression. Kc 

[NO]2 [Cl2 ] [NOCl]2



(0.4668) 2 (0.2334)

 0.0353

(1.200) 2

15.27

The quantity obtained by substituting initial concentrations into the equilibrium constant expression is called the reaction quotient (Qc). It differs from the equilibrium constant because initial concentrations rather than equilibrium concentrations are used.

15.28

1. 2. 3.

15.29

Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown quantity x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species.

Given: KP 

2 PSO

3

2 PSO PO2

 5.60  104

2

Initially, the total pressure is (0.350  0.762) atm or 1.112 atm. As the reaction progresses from left to right toward equilibrium there will be a decrease in the number of moles of molecules present. (Note that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3, or, at constant pressure, three atmospheres of reactants forms two atmospheres of products.) Since pressure is directly proportional to the number of molecules present, at equilibrium the total pressure will be less than 1.112 atm. 15.30

Strategy: We are given the initial concentrations of the gases, so we can calculate the reaction quotient (Qc). How does a comparison of Qc with Kc enable us to determine if the system is at equilibrium or, if not, in which direction the net reaction will proceed to reach equilibrium? Solution: Recall that for a system to be at equilibrium, Qc  Kc. Substitute the given concentrations into the equation for the reaction quotient to calculate Qc. Qc 

[NH3 ]02 [N 2 ]0 [H 2 ]30



[0.48]2 [0.60][0.76]3

 0.87

Comparing Qc to Kc, we find that Qc < Kc (0.87 < 1.2). The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds from left to right (consuming reactants, forming products) to reach equilibrium. Therefore, [NH3] will increase and [N2] and [H2] will decrease at equilibrium. 15.31

The balanced equation shows that one mole of carbon monoxide will combine with one mole of water to form hydrogen and carbon dioxide. Let x be the depletion in the concentration of either CO or H2O at equilibrium (why can x serve to represent either quantity?). The equilibrium concentration of hydrogen must then also be equal to x. The changes are summarized as shown in the table Initial (M): Change (M): Equilibrium (M):

H2  CO2 0 0 x x x x

ƒ H2O

 CO 0.0300 0.0300 x x (0.0300  x) (0.0300  x)

CHAPTER 15: CHEMICAL EQUILIBRIUM

Kc 

The equilibrium constant is:

[H 2 O][CO]  0.534 [H 2 ][CO2 ]

(0.0300  x) 2

Substituting,

x2

Taking the square root of both sides, we obtain:

357

 0.534

(0.0300  x)  x

0.534  0.731

x  0.0173 M The number of moles of H2 formed is: 0.0173 mol/L  10.0 L  0.173 mol H2 15.32

Strategy: The equilibrium constant KP is given, and we start with pure NO2. The partial pressure of O2 at equilibrium is 0.25 atm. From the stoichiometry of the reaction, we can determine the partial pressure of NO at equilibrium. Knowing KP and the partial pressures of both O2 and NO, we can solve for the partial pressure of NO2. Solution: Since the reaction started with only pure NO2, the equilibrium concentration of NO must be twice the equilibrium concentration of O2, due to the 2:1 mole ratio of the balanced equation. Therefore, the equilibrium partial pressure of NO is (2  0.25 atm)  0.50 atm. We can find the equilibrium NO2 pressure by rearranging the equilibrium constant expression, then substituting in the known values. KP 

2 PNO PO2

PNO2 

15.33

2 PNO

2

2 PNO PO2

KP



(0.50)2 (0.25)  0.020 atm 158

Notice that the balanced equation requires that for every two moles of HBr consumed, one mole of H2 and one mole of Br2 must be formed. Let 2x be the depletion in the concentration of HBr at equilibrium. The equilibrium concentrations of H2 and Br2 must therefore each be x. (Why?) The changes are shown in the table H2  Br2 ƒ 2HBr Initial (M): 0 0 0.267 Change (M): x x 2x Equilibrium (M): x x (0.267  2x) The equilibrium constant relationship is given by: Kc 

[HBr]2 [H 2 ][Br2 ]

358

CHAPTER 15: CHEMICAL EQUILIBRIUM

Substitution of the equilibrium concentration expressions gives Kc 

(0.267  2 x) 2 x

2

 2.18  106

Taking the square root of both sides we obtain: 0.267  2 x  1.48  103 x

x  1.80  10

4

The equilibrium concentrations are: [H2]  [Br2]  1.80  10

4

M 4

[HBr]  0.267  2(1.80  10 )  0.267 M If the depletion in the concentration of HBr at equilibrium were defined as x, rather than 2x, what would be the appropriate expressions for the equilibrium concentrations of H2 and Br2? Should the final answers be different in this case? 15.34

Strategy: We are given the initial amount of I2 (in moles) in a vessel of known volume (in liters), so we can calculate its molar concentration. Because initially no I atoms are present, the system could not be at equilibrium. Therefore, some I2 will dissociate to form I atoms until equilibrium is established. Solution: We follow the procedure outlined in Section 15.3 of the text to calculate the equilibrium concentrations. Step 1: The initial concentration of I2 is 0.0456 mol/2.30 L  0.0198 M. The stoichiometry of the problem shows 1 mole of I2 dissociating to 2 moles of I atoms. Let x be the amount (in mol/L) of I2 dissociated. It follows that the equilibrium concentration of I atoms must be 2x. We summarize the changes in concentrations as follows: I2(g) 0.0198 x (0.0198  x)

Initial (M): Change (M): Equilibrium (M):

ƒ

2I(g) 0.000 2x 2x

Step 2: Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Kc 

[I]2 (2 x)2   3.80  105 [I2 ] (0.0198  x) 5

2

7

4x  (3.80  10 )x  (7.52  10 )  0 2

The above equation is a quadratic equation of the form ax  bx  c  0. The solution for a quadratic equation is x 

b 

b 2  4ac 2a

CHAPTER 15: CHEMICAL EQUILIBRIUM

5

7

Here, we have a  4, b  3.80  10 , and c  7.52  10 . Substituting into the above equation, (3.80  105 ) 

x 

(3.80  105 )2  4(4)(  7.52  107 ) 2(4)

(3.80  105 )  (3.47  103 ) 8

x 

x  4.29  10

4

M

x  4.39  10

or

4

M

The second solution is physically impossible because you cannot have a negative concentration. The first solution is the correct answer. Step 3: Having solved for x, calculate the equilibrium concentrations of all species. [I]  2x  (2)(4.29  10

4

M)  8.58  10

4

M

4

[I2]  (0.0198  x)  [0.0198  (4.29  10 )] M  0.0194 M Tip: We could have simplified this problem by assuming that x was small compared to 0.0198. We could then assume that 0.0198  x  0.0198. By making this assumption, we could have avoided solving a quadratic equation. 15.35

Since equilibrium pressures are desired, we calculate KP. KP  Kc(0.0821 T)

n

COCl2(g) ƒ 0.760 x (0.760  x)

Initial (atm): Change (atm): Equilibrium (atm):

3

CO(g)  Cl2(g) 0.000 0.000 x x x x

x2  0.304 (0.760  x) 2

x  0.304x  0.231  0 x  0.352 atm At equilibrium:

PCOCl 2  (0.760  0.352)atm  0.408 atm PCO  0.352 atm

PCl 2  0.352 atm 15.36

(a)

The equilibrium constant, Kc, can be found by simple substitution. Kc 

1

 (4.63  10 )(0.0821  800)  0.304

[H 2 O][CO] (0.040)(0.050)   0.52 [CO2 ][H 2 ] (0.086)(0.045)

359

360

CHAPTER 15: CHEMICAL EQUILIBRIUM

(b)

The magnitude of the reaction quotient Qc for the system after the concentration of CO2 becomes 0.50 mol/L, but before equilibrium is reestablished, is: (0.040)(0.050)  0.089 (0.50)(0.045)

Qc 

The value of Qc is smaller than Kc; therefore, the system will shift to the right, increasing the concentrations of CO and H2O and decreasing the concentrations of CO2 and H2. Let x be the depletion in the concentration of CO2 at equilibrium. The stoichiometry of the balanced equation then requires that the decrease in the concentration of H2 must also be x, and that the concentration increases of CO and H2O be equal to x as well. The changes in the original concentrations are shown in the table. CO2  H2 ƒ CO  0.50 0.045 0.050 x x x (0.50  x) (0.045  x) (0.050  x)

Initial (M): Change (M): Equilibrium (M):

H2O 0.040 x (0.040  x)

The equilibrium constant expression is: [H 2 O][CO] (0.040  x )(0.050  x)   0.52 [CO 2 ][H 2 ] (0.50  x)(0.045  x)

Kc 

2

2

0.52(x  0.545x  0.0225)  x  0.090x  0.0020 3

2

0.48x  0.373x  (9.7  10 )  0 The positive root of the equation is x  0.025. The equilibrium concentrations are: [CO2]  (0.50  0.025) M  0.48 M [H2]  (0.045  0.025) M  0.020 M [CO]  (0.050  0.025) M  0.075 M [H2O]  (0.040  0.025) M  0.065 M 15.37

The equilibrium constant expression for the system is: KP 

( PCO ) 2 PCO2

The total pressure can be expressed as:

Ptotal  PCO2  PCO If we let the partial pressure of CO be x, then the partial pressure of CO2 is:

PCO2  Ptotal  x  (4.50  x)atm Substitution gives the equation: KP 

( PCO )2 x2   1.52 PCO2 (4.50  x)

CHAPTER 15: CHEMICAL EQUILIBRIUM

361

This can be rearranged to the quadratic: 2

x  1.52x  6.84  0 The solutions are x  1.96 and x  3.48; only the positive result has physical significance (why?). The equilibrium pressures are PCO  x  1.96 atm

PCO2  (4.50  1.96)  2.54 atm 15.38

The initial concentrations are [H2]  0.80 mol/5.0 L  0.16 M and [CO2]  0.80 mol/5.0 L  0.16 M. H2(g)  Initial (M): 0.16 Change (M): x Equilibrium (M): 0.16  x Kc 

CO2(g) ƒ 0.16 x 0.16  x

H2O(g)  CO(g) 0.00 0.00 x x x x

[H 2 O][CO] x2  4.2  [H 2 ][CO2 ] (0.16  x) 2

Taking the square root of both sides, we obtain: x  0.16  x

4.2

x  0.11 M The equilibrium concentrations are: [H2]  [CO2]  (0.16  0.11) M  0.05 M [H2O]  [CO]  0.11 M 15.39

Le Châtelier’s principle states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as it tries to reestablish equilibrium. The yield of a reaction can be maximized by changing concentrations of reactants or products, changing the temperature, or changing the pressure.

15.40

Vaporization is an endothermic reaction. Endothermic reactions are favored by an increase in temperature (the system shifts to the right). More gas is produced and hence the vapor pressure increases.

15.41

Changes in concentration, temperature, pressure, and volume can shift the position of an equilibrium. Changing the temperature can alter the value of the equilibrium constant.

15.42

The phrase “position of an equilibrium” refers to whether products or reactants are favored when a system reaches equilibrium. Changes in experimental conditions may shift the position of an equilibrium. When we say that an equilibrium position shifts to the right, we mean that the net reaction is now from left to right (reactants to products). A catalyst does not have any effect on the position of an equilibrium.

15.43

(a) (b) (c)

Addition of more Cl2(g) (a reactant) would shift the position of equilibrium to the right. Removal of SO2Cl2(g) (a product) would shift the position of equilibrium to the right. Removal of SO2(g) (a reactant) would shift the position of equilibrium to the left.

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CHAPTER 15: CHEMICAL EQUILIBRIUM

15.44

(a)

Removal of CO2(g) from the system would shift the position of equilibrium to the right.

(b)

Addition of more solid Na2CO3 would have no effect. [Na2CO3] does not appear in the equilibrium constant expression.

(c)

Removal of some of the solid NaHCO3 would have no effect. Same reason as (b).

(a)

This reaction is endothermic. (Why?) According to Section 15.4 of the text, an increase in temperature favors an endothermic reaction, so the equilibrium constant should become larger.

(b)

This reaction is exothermic. Such reactions are favored by decreases in temperature. The magnitude of Kc should decrease.

(c)

In this system heat is neither absorbed nor released. A change in temperature should have no effect on the magnitude of the equilibrium constant.

15.45

15.46

Strategy: A change in pressure can affect only the volume of a gas, but not that of a solid or liquid because solids and liquids are much less compressible. The stress applied is an increase in pressure. According to Le Châtelier's principle, the system will adjust to partially offset this stress. In other words, the system will adjust to decrease the pressure. This can be achieved by shifting to the side of the equation that has fewer moles of gas. Recall that pressure is directly proportional to moles of gas: PV  nRT so P  n. Solution: (a) Changes in pressure ordinarily do not affect the concentrations of reacting species in condensed phases because liquids and solids are virtually incompressible. Pressure change should have no effect on this system.

15.47

(b)

Same situation as (a).

(c)

Only the product is in the gas phase. An increase in pressure should favor the reaction that decreases the total number of moles of gas. The equilibrium should shift to the left, that is, the amount of B should decrease and that of A should increase.

(d)

In this equation there are equal moles of gaseous reactants and products. A shift in either direction will have no effect on the total number of moles of gas present. There will be no change when the pressure is increased.

(e)

A shift in the direction of the reverse reaction (left) will have the result of decreasing the total number of moles of gas present.

(a)

A pressure increase will favor the reaction (forward or reverse?) that decreases the total number of moles of gas. The equilibrium should shift to the right, i.e., more I2 will be produced at the expense of I.

(b)

If the concentration of I2 is suddenly altered, the system is no longer at equilibrium. Evaluating the magnitude of the reaction quotient Qc allows us to predict the direction of the resulting equilibrium shift. The reaction quotient for this system is: Qc 

[I 2 ]0 [I]02

Increasing the concentration of I2 will increase Qc. The equilibrium will be reestablished in such a way that Qc is again equal to the equilibrium constant. More I will form. The system shifts to the left to establish equilibrium. (c)

The forward reaction is exothermic. A decrease in temperature will shift the system to the right to reestablish equilibrium.

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.48

363

Strategy: (a) What does the sign of H indicate about the heat change (endothermic or exothermic) for the forward reaction? (b) The stress is the addition of Cl2 gas. How will the system adjust to partially offset the stress? (c) The stress is the removal of PCl3 gas. How will the system adjust to partially offset the stress? (d) The stress is an increase in pressure. The system will adjust to decrease the pressure. Remember, pressure is directly proportional to moles of gas. (e) What is the function of a catalyst? How does it affect a reacting system not at equilibrium? at equilibrium? Solution: (a) The stress applied is the heat added to the system. Note that the reaction is endothermic (H > 0). Endothermic reactions absorb heat from the surroundings; therefore, we can think of heat as a reactant. heat  PCl5(g) ƒ PCl3(g)  Cl2(g) The system will adjust to remove some of the added heat by undergoing a decomposition reaction (from left to right)

15.49

(b)

The stress is the addition of Cl2 gas. The system will shift in the direction to remove some of the added Cl2. The system shifts from right to left until equilibrium is reestablished.

(c)

The stress is the removal of PCl3 gas. The system will shift to replace some of the PCl3 that was removed. The system shifts from left to right until equilibrium is reestablished.

(d)

The stress applied is an increase in pressure. The system will adjust to remove the stress by decreasing the pressure. Recall that pressure is directly proportional to the number of moles of gas. In the balanced equation we see 1 mole of gas on the reactants side and 2 moles of gas on the products side. The pressure can be decreased by shifting to the side with the fewer moles of gas. The system will shift from right to left to reestablish equilibrium.

(e)

The function of a catalyst is to increase the rate of a reaction. If a catalyst is added to the reacting system not at equilibrium, the system will reach equilibrium faster than if left undisturbed. If a system is already at equilibrium, as in this case, the addition of a catalyst will not affect either the concentrations of reactant and product, or the equilibrium constant.

(a)

Increasing the temperature favors the endothermic reaction so that the concentrations of SO2 and O2 will increase while that of SO3 will decrease.

(b)

Increasing the pressure favors the reaction that decreases the number of moles of gas. The concentration of SO3 will increase.

(c)

Increasing the concentration of SO2 will lead to an increase in the concentration of SO3 and a decrease in the concentration of O2.

(d)

A catalyst has no effect on the position of equilibrium.

(e)

Adding an inert gas at constant volume has no effect on the position of equilibrium.

15.50

There will be no change in the pressures. A catalyst has no effect on the position of the equilibrium.

15.51

(a)

If helium gas is added to the system without changing the pressure or the temperature, the volume of the container must necessarily be increased. This will decrease the partial pressures of all the reactants and products. A pressure decrease will favor the reaction that increases the number of moles of gas. The position of equilibrium will shift to the left.

(b)

If the volume remains unchanged, the partial pressures of all the reactants and products will remain the same. The reaction quotient, Qc, will still equal the equilibrium constant, and there will be no change in the position of equilibrium.

364

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.52

For this system, KP  [CO2]. This means that to remain at equilibrium, the pressure of carbon dioxide must stay at a fixed value as long as the temperature remains the same. (a)

If the volume is increased, the pressure of CO2 will drop (Boyle's law, pressure and volume are inversely proportional). Some CaCO3 will break down to form more CO2 and CaO. (Shift right)

(b)

Assuming that the amount of added solid CaO is not so large that the volume of the system is altered significantly, there should be no change at all. If a huge amount of CaO were added, this would have the effect of reducing the volume of the container. What would happen then?

(c)

Assuming that the amount of CaCO3 removed doesn't alter the container volume significantly, there should be no change. Removing a huge amount of CaCO3 will have the effect of increasing the container volume. The result in that case will be the same as in part (a).

(d)

The pressure of CO2 will be greater and will exceed the value of KP. Some CO2 will combine with CaO to form more CaCO3. (Shift left)

(e)

Carbon dioxide combines with aqueous NaOH according to the equation CO2(g)  NaOH(aq)  NaHCO3(aq) This will have the effect of reducing the CO2 pressure and causing more CaCO3 to break down to CO2 and CaO. (Shift right)

(f)

Carbon dioxide does not react with hydrochloric acid, but CaCO3 does. CaCO3(s)  2HCl(aq)  CaCl2(aq)  CO2(g)  H2O(l) The CO2 produced by the action of the acid will combine with CaO as discussed in (d) above. (Shift left)

(g)

This is a decomposition reaction. Decomposition reactions are endothermic. Increasing the temperature will favor this reaction and produce more CO2 and CaO. (Shift right)

15.53

(i) (ii) (iii) (iv)

The temperature of the system is not given. It is not stated whether the equilibrium constant is KP or Kc (would they be different for this reaction?). A balanced equation is not given. The phases of the reactants and products are not given.

15.54

(a)

Since the total pressure is 1.00 atm, the sum of the partial pressures of NO and Cl2 is 1.00 atm  partial pressure of NOCl  1.00 atm  0.64 atm  0.36 atm The stoichiometry of the reaction requires that the partial pressure of NO be twice that of Cl2. Hence, the partial pressure of NO is 0.24 atm and the partial pressure of Cl2 is 0.12 atm.

(b)

The equilibrium constant KP is found by substituting the partial pressures calculated in part (a) into the equilibrium constant expression. KP 

15.55

2 PNO PCl2 2 PNOCl



(0.24) 2 (0.12) (0.64) 2

 0.017

Since KP increases with temperature, it is an endothermic reaction.

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.56

365

The equilibrium expression for this system is given by:

K P  PCO2 PH2O

15.57

(a)

In a closed vessel the decomposition will stop when the product of the partial pressures of CO2 and H2O equals KP. Adding more sodium bicarbonate will have no effect.

(b)

In an open vessel, CO2(g) and H2O(g) will escape from the vessel, and the partial pressures of CO2 and H2O will never become large enough for their product to equal KP. Therefore, equilibrium will never be established. Adding more sodium bicarbonate will result in the production of more CO2 and H2O.

The relevant relationships are: Kc 

[B]2 [A]

KP 

and

KP  Kc(0.0821 T)

n

PB2 PA

 Kc(0.0821 T)

n  1

We set up a table for the calculated values of Kc and KP. T (C)

KP

Kc 2

200

(0.843)  56.9 (0.0125)

56.9(0.0821  473)  2.21  10

3

300

(0.764)2  3.41 (0.171)

3.41(0.0821  573)  1.60  10

2

400

(0.724)2  2.10 (0.250)

2.10(0.0821  673)  116

Since Kc (and KP) decrease with temperature, the reaction is exothermic. 15.58

(a)

The equation that relates KP and Kc is: KP  Kc(0.0821 T)

n

For this reaction, n  3  2  1 Kc 

(b)

15.59

KP 2  1042   8  1044 (0.0821T ) (0.0821  298)

Because of a very large activation energy, the reaction of hydrogen with oxygen is infinitely slow without a catalyst or an initiator. The action of a single spark on a mixture of these gases results in the explosive formation of water.

Using data from Appendix 2 we calculate the enthalpy change for the reaction. H   2H f (NOCl)  2H f (NO)  H f (Cl2 )  2(51.7 kJ/mol)  2(90.4 kJ/mol)  (0)   77.4 kJ/mol

The enthalpy change is negative, so the reaction is exothermic. The formation of NOCl will be favored by low temperature.

366

CHAPTER 15: CHEMICAL EQUILIBRIUM

A pressure increase favors the reaction forming fewer moles of gas. The formation of NOCl will be favored by high pressure. 15.60

(a)

Calculate the value of KP by substituting the equilibrium partial pressures into the equilibrium constant expression. P (0.60) KP  B   1.7 2 (0.60) 2 PA

(b)

The total pressure is the sum of the partial pressures for the two gaseous components, A and B. We can write: PA  PB  1.5 atm and PB  1.5  PA Substituting into the expression for KP gives: KP 

(1.5  PA ) PA2

 1.7

1.7 PA2  PA  1.5  0

Solving the quadratic equation, we obtain: and by difference,

PA  0.69 atm PB  0.81 atm

Check that substituting these equilibrium concentrations into the equilibrium constant expression gives the equilibrium constant calculated in part (a). KP 

15.61

(a)

PB PA2



0.81 (0.69) 2

 1.7

The balanced equation shows that equal amounts of ammonia and hydrogen sulfide are formed in this decomposition. The partial pressures of these gases must just be half the total pressure, i.e., 0.355 atm. The value of KP is

K P  PNH3 PH2S  (0.355)2  0.126 (b)

We find the number of moles of ammonia (or hydrogen sulfide) and ammonium hydrogen sulfide. nNH3 

PV (0.355 atm)(4.000 L)   0.0582 mol RT (0.0821 L  atm/K  mol)(297 K)

nNH 4HS  6.1589 g 

1 mol  0.1205 mol (before decomposition) 51.12 g

From the balanced equation the percent decomposed is 0.0582 mol  100%  48.3% 0.1205 mol

CHAPTER 15: CHEMICAL EQUILIBRIUM

(c)

If the temperature does not change, KP has the same value. The total pressure will still be 0.709 atm at equilibrium. In other words the amounts of ammonia and hydrogen sulfide will be twice as great, and the amount of solid ammonium hydrogen sulfide will be: [0.1205  2(0.0582)]mol  0.0041 mol NH4HS

15.62

Total number of moles of gas is: 0.020  0.040  0.96  1.02 mol of gas You can calculate the partial pressure of each gaseous component from the mole fraction and the total pressure. 0.040 PNO   NO PT   0.20  0.0078 atm 1.02 PO2   O 2 PT 

0.020  0.20  0.0039 atm 1.02

PNO2   NO 2 PT 

0.96  0.20  0.19 atm 1.02

Calculate KP by substituting the partial pressures into the equilibrium constant expression. KP 

15.63

2 PNO

2

2 PNO PO2



(0.19) 2 2

(0.0078) (0.0039)

 1.5  105

Since the reactant is a solid, we can write:

K P  ( PNH3 ) 2 PCO2 The total pressure is the sum of the ammonia and carbon dioxide pressures.

Ptotal  PNH3  PCO2 From the stoichiometry,

PNH3  2 PCO2 Therefore:

Ptotal  2 PCO2  PCO2  3PCO2  0.318 atm PCO2  0.106 atm PNH3  0.212 atm Substituting into the equilibrium expression: 2

KP  (0.212) (0.106)  4.76  10 15.64

367

3

Set up a table that contains the initial concentrations, the change in concentrations, and the equilibrium concentration. Assume that the vessel has a volume of 1 L. Initial (M): Change (M): Equilibrium (M):

H2  0.47 x (0.47  x)

Cl2 0 x x

ƒ

2HCl 3.59 2x (3.59  2x)

368

CHAPTER 15: CHEMICAL EQUILIBRIUM

Substitute the equilibrium concentrations into the equilibrium constant expression, then solve for x. Since n  0, Kc  KP. Kc 

[HCl]2 (3.59  2 x) 2   193 [H 2 ][Cl2 ] (0.47  x) x

Solving the quadratic equation, x  0.10 Having solved for x, calculate the equilibrium concentrations of all species. [H2]  0.57 M

[Cl2]  0.10 M

[HCl]  3.39 M

Since we assumed that the vessel had a volume of 1 L, the above molarities also correspond to the number of moles of each component. From the mole fraction of each component and the total pressure, we can calculate the partial pressure of each component. Total number of moles  0.57  0.10  3.39  4.06 mol

15.65

(a)

PH 2 

0.57  2.00  0.28 atm 4.06

PCl 2 

0.10  2.00  0.049 atm 4.06

PHCl 

3.39  2.00  1.67 atm 4.06

From the balanced equation Initial (mol): Change (mol): Equilibrium (mol):

N2O4 ƒ 2NO2 1 0  2 (1  ) 2

The total moles in the system  (moles N2O4  moles NO2)  [(1  )  2]  1  . If the total pressure in the system is P, then: PN 2O4 

1  P and 1 

PNO2  2

KP 

2 PNO 2

PN 2O4

 2  2   P 1    1     P 1   

 4 2   P 1    2    4 P KP  1  1  2

2 P 1 

CHAPTER 15: CHEMICAL EQUILIBRIUM

(b)

369

Rearranging the KP expression: 2

2

4 P  KP   KP 2

 (4P  KP)  KP 2 

 

KP 4P  K P KP 4P  K P

KP is a constant (at constant temperature). Thus, as P increases,  must decrease, indicating that the system shifts to the left. This is also what one would predict based on the Chatelier's principle. 15.66

This is a difficult problem. Express the equilibrium number of moles in terms of the initial moles and the change in number of moles (x). Next, calculate the mole fraction of each component. Using the mole fraction, you should come up with a relationship between partial pressure and total pressure for each component. Substitute the partial pressures into the equilibrium constant expression to solve for the total pressure, PT. The reaction is: Initial (mol): Change (mol): Equilibrium (mol): Mole fraction of NH3 

 NH3  0.21 

 N2 1 x (1  x)

3 H2 ƒ 3 3x (3  3x)

2 NH3 0 2x 2x

mol of NH3 total number of moles

2x 2x  (1  x)  (3  3x)  2 x 4  2x

2x 4  2x

x  0.35 mol Substituting x into the following mole fraction equations, the mole fractions of N2 and H2 can be calculated.

 N2 

1 x 1  0.35   0.20 4  2x 4  2(0.35)

 H2 

3  3x 3  3(0.35)   0.59 4  2x 4  2(0.35)

The partial pressures of each component are equal to the mole fraction multiplied by the total pressure.

PNH3  0.21PT

PN2  0.20 PT

PH2  0.59 PT

Substitute the partial pressures above (in terms of PT) into the equilibrium constant expression, and solve for PT. KP 

2 PNH

3

PH3 PN 2 2

370

CHAPTER 15: CHEMICAL EQUILIBRIUM

4.31  104  4.31  104 

(0.21)2 PT2 (0.59 PT )3 (0.20 PT ) 1.07 PT 2 1

PT  5.0  10 atm

15.67

Kc 

For the balanced equation:

[H 2 ]2 [S2 ] [H 2S]2 2

 4.84  103  [S 2 ]  Kc    (2.25  104 )  2.34  103 M 2 3   [H 2 ]  1.50  10  [H 2S]2

15.68

We carry an additional significant figure throughout this calculation to minimize rounding errors. The initial molarity of SO2Cl2 is:

[SO2 Cl2 ] 

1 mol SO2 Cl2 135.0 g SO2 Cl2  0.02500 M 2.00 L

6.75 g SO 2 Cl2 

The concentration of SO2 at equilibrium is: [SO2 ] 

0.0345 mol  0.01725 M 2.00 L

Since there is a 1:1 mole ratio between SO2 and SO2Cl2, the concentration of SO2 at equilibrium (0.01725 M) equals the concentration of SO2Cl2 reacted. The concentrations of SO2Cl2 and Cl2 at equilibrium are: Initial (M): Change (M): Equilibrium (M):

SO2Cl2(g) 0.02500 0.01725 0.00775

ƒ SO2(g) 0 0.01725 0.01725



Cl2(g) 0 0.01725 0.01725

Substitute the equilibrium concentrations into the equilibrium constant expression to calculate Kc. Kc 

15.69

[SO2 ][Cl2 ] (0.01725)(0.01725)   3.84  102 [SO 2 Cl2 ] (0.00775)

For a 100% yield, 2.00 moles of SO3 would be formed (why?). An 80% yield means 2.00 moles  (0.80)  1.60 moles SO3 is formed. The amount of SO2 remaining at equilibrium  (2.00  1.60) mol  0.40 mol The amount of O2 reacted 

1 1  (amount of SO2 reacted)  (  1.60) mol  0.80 mol 2 2

The amount of O2 remaining at equilibrium  (2.00  0.80) mol  1.20 mol Total moles at equilibrium  moles SO2  moles O2  moles SO3  (0.40  1.20  1.60 ) mol  3.20 mol

CHAPTER 15: CHEMICAL EQUILIBRIUM

PSO2  PO2  PSO3 

KP 

0.13 

371

0.40 Ptotal  0.125 Ptotal 3.20 1.20 Ptotal  0.375 Ptotal 3.20 1.60 Ptotal  0.500 Ptotal 3.20

PSO3 2 PSO2 2 PO2 (0.500 Ptotal )2 (0.125 Ptotal ) 2 (0.375 Ptotal )

Ptotal  328 atm 15.70

I2(g) ƒ 2I(g) Assuming 1 mole of I2 is present originally and  moles reacts, at equilibrium: [I2]  1  , [I]  2. The total number of moles present in the system  (1  )  2  1  . From Problem 15.65(a) in the text, we know that KP is equal to: KP 

4 2 1  2

(1)

P

If there were no dissociation, then the pressure would be:  1 mol   L  atm  (1473 K) 1.00 g    0.0821  K  253.8 g mol nRT    P    0.953 atm V 0.500 L observed pressure 1.51 atm 1    calculated pressure 0.953 atm 1

  0.584 Substituting in equation (1) above: KP 

4 2 1  2

P 

(4)(0.584)2 1  (0.584)2

 1.51  3.13

15.71

Panting decreases the concentration of CO2 because CO2 is exhaled during respiration. This decreases the concentration of carbonate ions, shifting the equilibrium to the left. Less CaCO3 is produced. Two possible solutions would be either to cool the chickens' environment or to feed them carbonated water.

15.72

According to the ideal gas law, pressure is directly proportional to the concentration of a gas in mol/L if the reaction is at constant volume and temperature. Therefore, pressure may be used as a concentration unit. The reaction is:  3H2 ƒ 2NH3 N2 Initial (atm): 0.862 0.373 0 Change (atm): x 3x 2x Equilibrium (atm): (0.862  x) (0.373  3x) 2x

372

CHAPTER 15: CHEMICAL EQUILIBRIUM

KP 

2 PNH

3

PH3 PN 2 2

4.31  104 

(2 x)2 (0.373  3 x)3 (0.862  x)

At this point, we need to make two assumptions that 3x is very small compared to 0.373 and that x is very small compared to 0.862. Hence, 0.373  3x  0.373 and 0.862  x  0.862 4.31  104 

Solving for x. x  2.20  10

(2 x)2 (0.373)3 (0.862) 3

atm

The equilibrium pressures are:

PN2  [0.862  (2.20  103 )]atm  0.860 atm PH 2  [0.373  (3)(2.20  103 )]atm  0.366 atm

PNH 3  (2)(2.20  103 atm)  4.40  103 atm Was the assumption valid that we made above? Typically, the assumption is considered valid if x is less than 5 percent of the number that we said it was very small compared to. Is this the case? 15.73

(a)

The sum of the mole fractions must equal one.

 CO   CO2  1

and

 CO2  1   CO

According to the hint, the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass. (CO  28.01 g)  [(1  CO)  44.01 g]  35 g Solving, (b)

Solving for the pressures

XCO  0.56

and

 CO2  0.44

Ptotal = PCO + PCO2  11 atm PCO  COPtotal  0.56  11 atm  6.2 atm

PCO2   CO2 Ptotal  (0.44)(11 atm)  4.8 atm KP 

2 PCO (6.2)2   8.0 PCO2 4.8

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.74

(a)

The equation is: Initial (M): Change (M): Equilibrium (M):

ƒ

fructose 0.244 0.131 0.113

373

glucose 0 0.131 0.131

Calculating the equilibrium constant, Kc 

(b)

Percent converted  

[glucose] 0.131   1.16 [fructose] 0.113

amount of fructose converted  100% original amount of fructose 0.131  100%  53.7% 0.244

15.75

If you started with radioactive iodine in the solid phase, then you should find radioactive iodine in the vapor phase at equilibrium. Conversely, if you started with radioactive iodine in the vapor phase, you should find radioactive iodine in the solid phase. Both of these observations indicate a dynamic equilibrium between solid and vapor phase.

15.76

(a)

There is only one gas phase component, O2. The equilibrium constant is simply

K P  PO2 so, KP  0.49 (b)

From the ideal gas equation, we can calculate the moles of O2 produced by the decomposition of CuO. nO2 

PV (0.49 atm)(2.0 L)   9.2  103 mol O2 (0.0821 L  atm/K  mol)(1297 K) RT

From the balanced equation, (9.2  103 mol O2 ) 

4 mol CuO  3.7  102 mol CuO decomposed 1 mol O2

Fraction of CuO decomposed 

(c)

amount of CuO lost 3.7  102 mol   0.23 original amount of CuO 0.16 mol

If a 1.0 mole sample were used, the pressure of oxygen would still be the same (0.49 atm) and it would be due to the same quantity of O2. Remember, a pure solid does not affect the equilibrium position. 2 The moles of CuO lost would still be 3.7  10 mol. Thus the fraction decomposed would be: 0.037  0.037 1.0

(d)

2

If the number of moles of CuO were less than 3.7  10 mol, the equilibrium could not be established because the pressure of O2 would be less than 0.49 atm. Therefore, the smallest number of moles of 2 CuO needed to establish equilibrium must be slightly greater than 3.7  10 mol.

374

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.77

If there were 0.88 mole of CO2 initially and at equilibrium there were 0.11 mole, then (0.88  0.11) mole  0.77 mole reacted. NO  CO2 ƒ NO2  CO Initial (mol): 3.9 0.88 0 0 Change (mol): 0.77 0.77 0.77 0.77 Equilibrium (mol): (3.9  0.77) 0.11 0.77 0.77 Solving for the equilibrium constant:

Kc 

(0.77)(0.77)  1.7 (3.9  0.77)(0.11)

In the balanced equation there are equal number of moles of products and reactants; therefore, the volume of the container will not affect the calculation of Kc. We can solve for the equilibrium constant in terms of moles. 15.78

We first must find the initial concentrations of all the species in the system. [H 2 ]0 

0.714 mol  0.298 M 2.40 L

[I2 ]0 

0.984 mol  0.410 M 2.40 L

[HI]0 

0.886 mol  0.369 M 2.40 L

Calculate the reaction quotient by substituting the initial concentrations into the appropriate equation. [HI]02 (0.369)2   1.11 [H 2 ]0 [I 2 ]0 (0.298)(0.410)

Qc 

We find that Qc is less than Kc. The equilibrium will shift to the right, decreasing the concentrations of H2 and I2 and increasing the concentration of HI. We set up the usual table. Let x be the decrease in concentration of H2 and I2.  H2 0.298 x (0.298  x)

Initial (M): Change (M): Equilibrium (M):

I2 ƒ 0.410 x (0.410  x)

2 HI 0.369 2x (0.369  2x)

The equilibrium constant expression is: Kc 

[HI]2 (0.369  2 x )2   54.3 [H 2 ][I 2 ] (0.298  x)(0.410  x)

This becomes the quadratic equation 2

50.3x  39.9x  6.48  0 The smaller root is x  0.228 M. (The larger root is physically impossible.)

CHAPTER 15: CHEMICAL EQUILIBRIUM

375

Having solved for x, calculate the equilibrium concentrations. [H2]  (0.298  0.228) M  0.070 M [I2]  (0.410  0.228) M  0.182 M [HI]  [0.369  2(0.228)] M  0.825 M 15.79

Since we started with pure A, then any A that is lost forms equal amounts of B and C. Since the total pressure is P, the pressure of B  C  P  0.14 P  0.86 P. The pressure of B  C  0.43 P. PB PC (0.43 P )(0.43 P )   1.3 P PA 0.14 P

KP 

15.80

The gas cannot be (a) because the color became lighter with heating. Heating (a) to 150C would produce some HBr, which is colorless and would lighten rather than darken the gas. The gas cannot be (b) because Br2 doesn't dissociate into Br atoms at 150C, so the color shouldn't change. The gas must be (c). From 25C to 150C, heating causes N2O4 to dissociate into NO2, thus darkening the color (NO2 is a brown gas). N2O4(g)  2NO2(g) Above 150C, the NO2 breaks up into colorless NO and O2. 2NO2(g)  2NO(g)  O2(g) An increase in pressure shifts the equilibrium back to the left, forming NO2, thus darkening the gas again. 2NO(g)  O2(g)  2NO2(g)

15.81

Given the following:

(a)

Kc 

[NH3 ]2

 0.65

[N 2 ][H 2 ]3

Temperature must have units of Kelvin. KP  Kc(0.0821 T)

n

KP  (0.65)(0.0821  668) (b)

(24)

Recalling that, 1 K reverse

Kforward 

Therefore, K c' 

(c)

1  1.5 0.65

Since the equation 3 2

H2(g) ƒ NH3(g)

1 2

N2(g) 

1 2

[N2(g)  3H2(g) ƒ 2NH3(g)]

is equivalent to

 2.2  10

4

376

CHAPTER 15: CHEMICAL EQUILIBRIUM

then, Kc' for the reaction: 1 N (g)  2 2

3 2

H2(g) ƒ NH3(g)

1

equals ( Kc ) 2 for the reaction: N2(g)  3H2(g) ƒ 2NH3(g) Thus, 1

K c'  ( Kc ) 2 

(d)

0.65  0.81

For KP in part (b): KP  (1.5)(0.0821  668)

2

 4.5  10

3

and for KP in part (c): KP  (0.81)(0.0821  668) 15.82

 0.015

Decomposition of N2O4 is an endothermic process, N2O4 ƒ 2NO2. As the temperature is increased, the system shifts to the right to reestablish equilibrium. (a) (d)

15.83

1

Color deepens increases

(b) (e)

increases (c) decreases unchanged (conservation of mass)

The vapor pressure of water is equivalent to saying the partial pressure of H2O(g).

K P  PH 2O  0.0231 Kc 

KP (0.0821T )

n



0.0231 1

(0.0821  293)

 9.60  104

15.84

Potassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right.

15.85

This problem involves two types of problems: Calculations using density or molar mass and Dalton’s Law of Partial Pressures. We can calculate the average molar mass of the gaseous mixture from the density. M=

dRT P

Let M be the average molar mass of NO2 and N2O4. The above equation becomes: M 

dRT (2.9 g/L)(0.0821 L  atm/K  mol)(347 K)  P 1.3 atm

M  63.6 g/mol

The average molar mass is equal to the sum of the molar masses of each component times the respective mole fractions. Setting this up, we can calculate the mole fraction of each component.

CHAPTER 15: CHEMICAL EQUILIBRIUM

377

M   NO 2 M NO 2   N 2O 4 M N 2O 4  63.6 g/mol

 NO2 (46.01 g/mol)  (1   NO2 )(92.02 g/mol)  63.6 g/mol

 NO2  0.618 We can now calculate the partial pressure of NO2 from the mole fraction and the total pressure.

PNO2   NO2 PT

PNO2  (0.618)(1.3 atm)  0.80 atm We can calculate the partial pressure of N2O4 by difference.

PN2O4  PT  PNO2

PN2O4  (1.3  0.80) atm  0.50 atm Finally, we can calculate KP for the dissociation of N2O4. KP 

15.86

(a)

2 PNO

2

PN 2O4



(0.80)2  1.3 0.50

Since both reactions are endothermic (H is positive), according to Le Châtelier’s principle the products would be favored at high temperatures. Indeed, the steam-reforming process is carried out at very high temperatures (between 800C and 1000C). It is interesting to note that in a plant that uses natural gas (methane) for both hydrogen generation and heating, about one-third of the gas is burned to maintain the high temperatures. In each reaction there are more moles of products than reactants; therefore, we expect products to be favored at low pressures. In reality, the reactions are carried out at high pressures. The reason is that when the hydrogen gas produced is used captively (usually in the synthesis of ammonia), high pressure leads to higher yields of ammonia.

(b)

(i)

The relation between Kc and KP is given by Equation (15.5) of the text: KP  Kc(0.0821 T)

n

Since n  4  2  2, we write: 2

KP  (18)(0.0821  1073)  1.4  10 (ii)

5

Let x be the amount of CH4 and H2O (in atm) reacted. We write: Initial (atm): Change (atm): Equilibrium (atm):

CH4  H2O ƒ 15 15 x x 15  x 15  x

The equilibrium constant is given by: KP 

PCO PH3

2

PCH4 PH 2O

CO  3H2 0 0 x 3x x 3x

378

CHAPTER 15: CHEMICAL EQUILIBRIUM

1.4  105 

( x )(3 x)3 27 x 4  (15  x)(15  x ) (15  x )2

Taking the square root of both sides, we obtain: 3.7  102 

5.2 x 2 15  x

which can be expressed as 2

2

3

5.2x  (3.7  10 x)  (5.6  10 )  0 Solving the quadratic equation, we obtain x  13 atm (The other solution for x is negative and is physically impossible.) At equilibrium, the pressures are:

PCH4  (15  13)  2 atm PH 2O  (15  13)  2 atm PCO  13 atm

PH 2  3(13 atm)  39 atm shifts to right no change

15.87

(a) (e)

15.88

K P = PNH3 PHCl PNH3  PHCl 

(b) (f)

shifts to right shifts to left

(c) (g)

no change shifts to right

(d)

no change

2.2  1.1 atm 2

KP  (1.1)(1.1)  1.2 15.89

The equilibrium is:

N2O4(g) ƒ 2NO2(g) KP 

( PNO 2 ) 2 PN 2O4



0.152  0.113 0.20

Volume is doubled so pressure is halved. Let’s calculate QP and compare it to KP. 2

 0.15   2    0.0563  K QP   P  0.20   2   

Equilibrium will shift to the right. Some N2O4 will react, and some NO2 will be formed. Let’s let x  amount of N2O4 reacted.

CHAPTER 15: CHEMICAL EQUILIBRIUM

Initial (atm): Change (atm): Equilibrium (atm):

N2O4(g) ƒ 0.10 x 0.10  x

379

2NO2(g) 0.075 2x 0.075  2x

Substitute into the KP expression to solve for x. (0.075  2 x)2 0.10  x

K P  0.113  2

4x  0.413x  5.67  10

3

 0

x  0.0123 At equilibrium:

PNO2  0.075  2(0.0123)  0.0996  0.100 atm PN 2O4  0.10  0.0123  0.09 atm Check: KP 

15.90

(0.10)2  0.111 0.09

close enough to 0.113

(a)

React Ni with CO above 50C. Pump away the Ni(CO)4 vapor (shift equilibrium to right), leaving the solid impurities behind.

(b)

Consider the reverse reaction: Ni(CO)4(g)  Ni(s)  4CO(g) H   4H f (CO)  H f [Ni(CO)4 ]

H  (4)(110.5 kJ/mol)  (1)(602.9 kJ/mol)  160.9 kJ/mol The decomposition is endothermic, which is favored at high temperatures. Heat Ni(CO)4 above 200C to convert it back to Ni. 15.91

(a)

Molar mass of PCl5  208.2 g/mol  1 mol   L  atm   523 K   2.50 g    0.0821 208.2 g   mol  K  nRT  P    1.03 atm V 0.500 L

(b) Initial (atm) Change (atm) Equilibrium (atm)

PCl5 ƒ 1.03 x 1.03  x

K P  1.05 

x2 1.03  x

PCl3 0 x x



Cl2 0 x x

380

CHAPTER 15: CHEMICAL EQUILIBRIUM

2

x  1.05x  1.08  0 x  0.639 At equilibrium:

PPCl5  1.03  0.639  0.39 atm

15.92

(c)

PT  (1.03  x)  x  x  1.03  0.639  1.67 atm

(d)

0.639 atm  0.620 1.03 atm

(a)

KP  PHg  0.0020 mmHg  2.6  10 Kc 

(b)

KP (0.0821T )

n



6

atm  2.6  10

2.6  106

6

(equil. constants are expressed without units)

 1.1  107

1

(0.0821  299)

3

Volume of lab  (6.1 m)(5.3 m)(3.1 m)  100 m [Hg]  Kc

3

Total mass of Hg vapor 

 1 cm  1.1  107 mol 200.6 g 1L 3      100 m  2.2 g 3 1L 1 mol 1000 cm  0.01 m 

The concentration of mercury vapor in the room is: 2.2 g 100 m3

 0.022 g/m3  22 mg/m 3 3

Yes! This concentration exceeds the safety limit of 0.05 mg/m . Better clean up the spill! 15.93

15.94

(a)

A catalyst speeds up the rates of the forward and reverse reactions to the same extent.

(b)

A catalyst would not change the energies of the reactant and product.

(c)

The first reaction is exothermic. Raising the temperature would favor the reverse reaction, increasing the amount of reactant and decreasing the amount of product at equilibrium. The equilibrium constant, K, would decrease. The second reaction is endothermic. Raising the temperature would favor the forward reaction, increasing the amount of product and decreasing the amount of reactant at equilibrium. The equilibrium constant, K, would increase.

(d)

A catalyst lowers the activation energy for the forward and reverse reactions to the same extent. Adding a catalyst to a reaction mixture will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait longer for equilibrium to be reached. If the same equilibrium position is reached, with or without a catalyst, then the equilibrium constant is the same.

First, let's calculate the initial concentration of ammonia. 1 mol NH3 17.03 g NH3  0.214 M 4.00 L

14.6 g  [NH3 ] 

CHAPTER 15: CHEMICAL EQUILIBRIUM

381

Let's set up a table to represent the equilibrium concentrations. We represent the amount of NH3 that reacts as 2x. 2NH3(g) ƒ N2(g)  3H2(g) Initial (M): 0.214 0 0 Change (M): 2x x 3x Equilibrium (M): 0.214  2x x 3x Substitute into the equilibrium constant expression to solve for x. Kc 

[N 2 ][H 2 ]3

0.83 

[NH3 ]2 ( x)(3 x)3



(0.214  2 x)2

27 x 4 (0.214  2 x) 2

Taking the square root of both sides of the equation gives: 0.91 

Rearranging,

5.20 x 2 0.214  2 x

2

5.20x  1.82x  0.195  0 Solving the quadratic equation gives the solutions: x  0.086 M and x  0.44 M The positive root is the correct answer. The equilibrium concentrations are: [NH3]  0.214  2(0.086)  0.042 M [N2]  0.086 M [H2]  3(0.086)  0.26 M 15.95

(a)

The sum of the mole fractions must equal one.

 N 2O4   NO2  1

 N 2O4  1   NO2

and

The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.

( Χ NO2  46.01 g)  ( Χ N2O4  92.02 g)  70.6 g ( Χ NO2  46.01 g)  [(1  Χ NO2 )  92.02 g]  70.6 g Solving, (b)

 NO2  0.466

and

 N2O4  0.534

Solving for the pressures:

Ptotal = PN2O4 + PNO2  1.2 atm PNO2   NO2 Ptotal  (0.466)(1.2 atm)  0.56 atm PN 2O4   N2O4 Ptotal  (0.534)(1.2 atm)  0.64 atm

382

CHAPTER 15: CHEMICAL EQUILIBRIUM

KP 

(c)

2 PNO

2

PN 2O4



(0.56)2  0.49 0.64

The total pressure is increased to 4.0 atm. We know that,

PNO2   NO2 Ptotal PN 2O4   N2O4 Ptotal  (1  Χ NO2 ) Ptotal Using the KP value calculated in part (b), we can solve for the mole fraction of NO2. KP 

2 PNO

2

PN2O4

0.487 

0.487 



( X NO2 PT )2 (1  X NO2 ) PT

[ X NO2 (4.0)]2 (1  X NO2 )(4.0) 2 16 X NO

2

4.0  4.0 X NO2

2 1.95  1.95 X NO2  16 X NO 2 2 0  16 X NO  1.95 X NO2  1.95 2

Solving the quadratic equation,

 NO2  0.29  N2O4  1   NO2  0.71 As the pressure is increased, LeChâtelier’s principle tells us that the system will shift to reduce the pressure. The system will shift to the side of the equation with fewer moles of gas to reduce the pressure. As the mole fractions calculated show, the system shifted to increase the amount of N2O4 and decrease the amount of NO2. There are now fewer moles of gas in the system. You can check your work by calculating the new partial pressures of NO2 and N2O4, and then substitute the values into the KP expression to solve for KP. If the problem is solved correctly, you will obtain the KP value calculated in part (b). 15.96

The equilibrium constant expression for the given reaction is KP 

We also know that

2 PCO PCO2

PCO   CO Ptotal

and

PCO2   CO2 Ptotal

CHAPTER 15: CHEMICAL EQUILIBRIUM

383

We can solve for the mole fractions of each gas, and then substitute into the equilibrium constant expression to solve for the total pressure. We carry an additional significant figure throughout this calculation to minimize rounding errors.

 CO  and

0.025 mol  0.676 0.037 mol

 CO2  1   CO  0.324

Substitute into the equilibrium constant expression. KP 

2 PCO (  CO PT )2   CO2 PT PCO2

1.9 

(0.676 PT )2 0.324 PT

1.9  1.41PT PT  1.3 atm Check:

PCO   CO Ptotal  (0.676)(1.35 atm)  0.913 atm

PCO2   CO2 Ptotal  (0.324)(1.35 atm)  0.437 atm KP 

15.97

2 PCO (0.913)2   1.9 PCO2 0.437

Since the catalyst is exposed to the reacting system, it would catalyze the 2A  B reaction. This shift would result in a decrease in the number of gas molecules, so the gas pressure decreases. The piston would be pushed down by the atmospheric pressure. When the cover is over the box, the catalyst is no longer able to favor the forward reaction. To reestablish equilibrium, the B  2A step would dominate. This would increase the gas pressure so the piston rises and so on. Conclusion: Such a catalyst would result in a perpetual motion machine (the piston would move up and down forever) which can be used to do work without input of energy or net consumption of chemicals. Such a machine cannot exist.

15.98

Initially, at equilibrium: [NO2]  0.0475 M and [N2O4]  0.491 M. At the instant the volume is halved, the concentrations double. [NO2]  2(0.0475 M)  0.0950 M and [N2O4]  2(0.491 M)  0.982 M. The system is no longer at equilibrium. The system will shift to the left to offset the increase in pressure when the volume is halved. When a new equilibrium position is established, we write: N2O4 ƒ 0.982 M  x

2NO2 0.0950 M – 2x

K c  4.63  103  2

[NO2 ]2 (0.0950  2 x) 2  [N 2 O 4 ] (0.982  x)

4x – 0.3846x  4.478  10

3

0

384

CHAPTER 15: CHEMICAL EQUILIBRIUM

Solving

x  0.0826 M (impossible)

and x  0.0136 M

At the new equilibrium, [N2O4]  0.982  0.0136  0.996 M [NO2]  0.0950 – (2  0.0136)  0.0678 M As we can see, the new equilibrium concentration of NO2 is greater than the initial equilibrium concentration (0.0475 M). Therefore, the gases should look darker! 15.99

To determine H, we need to plot ln KP versus 1/T (y vs. x). 1/T 0.00167 0.00143 0.00125 0.00111 0.00100

ln KP 4.93 1.63 0.83 2.77 4.34 5 4 3 2

y = 1.38E+04x - 1.81E+01

ln KP

1 0 -1 -2 -3 -4 -5 0.0010

0.0011

0.0012

0.0013

0.0014

0.0015

0.0016

1

1/T (K )

The slope of the plot equals H/R. 1.38  104 K  

H  8.314 J/mol  K 5

H  1.15 × 10 J/mol  115 kJ/mol 15.100 (a) We start by writing the van’t Hoff equation at two different temperatures. ln K1 

H  C RT1

ln K 2 

H  C RT2

0.0017

CHAPTER 15: CHEMICAL EQUILIBRIUM

ln K1  ln K 2 

385

H  H   RT1 RT2

K H   1  ln 1     K2 R  T2 T1 

Assuming an endothermic reaction, H > 0 and T2 > T1. Then,

H   1    < 0, meaning that  R  T2 T1 

K ln 1 < 0 or K1 < K2. A larger K2 indicates that there are more products at equilibrium as the temperature is K2 raised. This agrees with LeChatelier’s principle that an increase in temperature favors the forward endothermic reaction. The opposite of the above discussion holds for an exothermic reaction.

(b) Treating H2O(l) ƒ H2O(g)

Hvap  ?

as a heterogeneous equilibrium, K P  PH2O . We substitute into the equation derived in part (a) to solve for Hvap. K H   1  ln 1     K2 R  T2 T1  ln

 1 31.82 mmHg H        92.51 mmHg 8.314 J/mol  K  323 K 303 K  5

1.067  H(2.458 × 10 ) 4

H  4.34 × 10 J/mol  43.4 kJ/mol 15.101 Using Equation (14.10) of the text, we can calculate k1. k  Ae  Ea /RT

Then, we can calculate k1 using the expression Kc 

k1 (see Section 15.1 of the text) k1

k1  (1.0

  41  103 J/mol    (8.314 J/mol K)(298 K)    1012 s 1 )e 

4 1

k1  6.5 × 10 s Kc 

k1 k1

9.83  103 

k1

6.5  104 s 1 8 1

k1  6.4 × 10 s

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CHAPTER 15: CHEMICAL EQUILIBRIUM

15.102 We start with a table.

Initial (mol): Change (mol): Equilibrium (mol):



A2 1 x  2 1

ƒ

B2 3 x  2

x 2

3

2AB 0 x

x 2

x

After the addition of 2 moles of A,

Initial (mol): Change (mol): Equilibrium (mol):

A2 x 3 2 x  2 3x



B2 ƒ x 3 2 x  2 3x

2AB x x 2x

We write two different equilibrium constants expressions for the two tables. K 

K 

[AB]2 [A 2 ][B2 ] x2 (2 x) 2 and K  x  x (3  x)(3  x)  1    3   2  2 

We equate the equilibrium constant expressions and solve for x. x2 (2 x)2  x  x (3  x)(3  x)  1   3   2 2    1 4  2 1 2 x  6x  9 ( x  8 x  12) 4

6x  9  8x  12 x  1.5 We substitute x back into one of the equilibrium constant expressions to solve for K. K 

(2 x) 2 (3)2   4.0 (3  x)(3  x) (1.5)(1.5)

Substitute x into the other equilibrium constant expression to see if you obtain the same value for K. Note that we used moles rather than molarity for the concentrations, because the volume, V, cancels in the equilibrium constant expressions.

CHAPTER 15: CHEMICAL EQUILIBRIUM

15.103 (a)

387

First, we calculate the moles of I2. mol I2  0.032 g I2 

1 mol I 2  1.26  104 mol 253.8 g I2 4

Let x be the number of moles of I2 that dissolves in CCl4, so (1.26 × 10  x)mol remains dissolved in water. We set up expressions for the concentrations of I2 in CCl4 and H2O. (1.26  104  x) mol x mol and [I2 (CCl4 )]  0.200 L 0.030 L

[I2 (aq )] 

Next, we substitute these concentrations into the equilibrium constant expression and solve for x. K 

[I 2 (CCl4 )] [I2 (aq )]

x 0.030 83  (1.26  104  x) 0.200 4

83(1.26 × 10

 x)  6.67x 4

x  1.166 × 10

The fraction of I2 remaining in the aqueous phase is: fraction ( f ) 

(1.26  104 )  (1.166  104 ) 1.26  104

 0.075

(b)

The first extraction leaves only 7.5% I2 in the water. The next extraction with 0.030 L of CCl4 will leave 3 only (0.075)(0.075)  5.6 × 10 . This is the fraction remaining after the second extraction which is only 0.56%.

(c)

For a single extraction using 0.060 L of CCl4, we let y be the number of moles of I2 in CCl4. y 0.060 83  (1.26  104  y ) 0.200 4

83(1.26 × 10

 y)  3.33y 4

y  1.211 × 10

The fraction of I2 remaining in the aqueous phase is: fraction ( f ) 

(1.26  104 )  (1.211  104 ) 1.26  104

 0.039

The fraction of I2 remaining dissolved in water is 0.039 or 3.9%. The extraction with 0.060 L of CCl4 is not as effective as two separate extractions of 0.030 L each.

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CHAPTER 15: CHEMICAL EQUILIBRIUM

15.104 (a)

As volume of a gas is increased, pressure decreases at constant temperature (see Figure 5.6 of the text). As pressure decreases, LeChâtelier’s principle tells us that a system will shift to increase the pressure by shifting to the side of the equation with more moles of gas. In this case, the system will shift to the right. As volume of the system is increased, more NO2 will be produced increasing the molecules of gas in the system, thereby increasing the pressure of the system compared to an ideal gas. Note that a large volume corresponds to a small value of 1/V. A plot of P versus 1/V at constant temperature for the system and an ideal gas is shown below. Ideal gas

system P

1/V (b)

As temperature of a gas is increased, the volume occupied by the gas increases at constant pressure (See Figure 5.8 of the text). Because this reaction is endothermic, the system will shift to the right as the temperature is increased. As the system shifts right, more NO2 will be produced increasing the molecules of gas in the system, thereby increasing the volume occupied by these gases compared to an ideal gas. A plot of V versus T at constant pressure for the system and an ideal gas is shown below. system Ideal gas V

T